Question: Evaluate the improper integral if it exists. $\int_{0}^{1}\dfrac{1}{(x-1)^{\frac13}}\,dx $ Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac32$ (Choice B) B $-1$ (Choice C) C $\dfrac23$ (Choice D) D The improper integral diverges.
Solution: First, let's rewrite the improper integral: $\int_{0}^{1}\dfrac{1}{(x-1)^{\frac13}}\,dx= \lim_{b\to 1^-}\int_0^b \dfrac{1}{(x-1)^{\frac13}}\,dx$ We can now evaluate the integral: $\begin{aligned} \phantom{\int_0^{1}\dfrac1{\sqrt{x}}dx}&=\lim_{b\to 1^-}\int_0^b \dfrac{1}{(x-1)^{\frac13}}\,dx\\ \\ \\ &=\lim_{b\to1^-} \left[\dfrac32(x-1)^{2/3}\right]_0^b\\ \\ \\ &=\dfrac32\lim_{b\to1^-}\left[(b-1)^{2/3}-(0-1)^{2/3}\right]\\ \\ \\ &=\dfrac32\left(\lim_{b\to1^-}(b-1)^{2/3}-\lim_{b\to1^-}1\right)\\ \\ &=\dfrac32(0-1)\\ \\ &=-\dfrac32 \end{aligned}$ The answer: $\int_{0}^{1}\dfrac{1}{(x-1)^{\frac13}}\,dx =-\dfrac32$